Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8

Question 1.

Verify Rolle’s Theorem for the function f(x) = xÂ² + 2x – 8, x âˆˆ [- 4, 2].

Solution:

Now, f(x) = xÂ² + 2x – 8 is a polynomial. Therefore, it is continuous and derivable in its domain x âˆˆ R. Hence, it is continuous in the interval [- 4, 2] and derivable in the interval (- 4, 2).

f(- 4) = (- 4)Â² + 2(- 4) – 8 = 16 – 8 – 8 = 0

and f(2) = 2Â² + 4 – 8 = 8 – 8 = 0.

âˆ´ Conditions of Rolle’s theorem are satisfied.

Now, f'(x) = 2x + 2

f'(c) = 2c + 2 = 0 or c = – 1.

Now, c = – 1 âˆˆ [- 4, 2]

Thus, f'(c) = 0 at c = – 1.

Question 2.

Examine if Rolle’s theorem is applicable to any of the following functions. Can you say something about the converse of Rolle’s theorem from these examples?

(i) f(x) = [x] for x âˆˆ [5, 9]

(ii) f(x) = [x] for x âˆˆ [- 2, 2]

(iii) f(x) = xÂ² – 1 for x âˆˆ [1, 2]

Solution:

(i) In the interval [5, 9], f(x) = [x] is neither continuous nor derivable at x = 6,7,8.

Hence, Rolle’s theorem is not applicable.

(ii) f(x) = [x] is not continuous and derivable at – 1, 0, 1.

Hence, Rolle’s theorem is not applicable.

(iii) f(x) = (xÂ² – 1), f(1) = 1 – 1 = 0, f(2) = 2Â² – 1 = 3

Thus, f(a) â‰ f(b).

Though it is continuous in the interval [1, 2] and derivable in the interval (1, 2), still Rolle’s Theorem is not applicable.

In case of converse, if f'(c) = 0, c âˆˆ [a, b], then conditions of Rolle’s theorem are not true.

(i) f(x) = [x] is the greatest integer less than or equal to x.

âˆ´ f’ (x) = 0. But f is neither continuous in the interval [5, 9] nor differentiable in the interval (5, 9).

(ii) Here also, though f'(x) = 0, but fis neither continuous in interval [- 2, 2] nor differentiable in the interval (- 2, 2).

(iii) f(x) = xÂ² – 1, f'(x) = 2x. Here f'(x) is not zero in the interval [1, 2]. So, f(2) â‰ f(1)

Question 3.

If f: [- 5, 5] â†’ R is a differentiable function and if f’ (x) does not vanish anywhere, then prove that f(- 5) â‰ f(5).

Solution:

For Rolle’s theorem: If

(i) f is continuous in [a, b],

(ii) f is derivable in (a, b).

(iii) f(a) = f(b)

then f'(c) = 0, c âˆˆ (a, b)

We are given f is continuous and det

but f'(c) â‰ 0 â‡’ f(a)

i.e., f(-5) â‰ f(5).

Question 4.

Verify Mean Value Theorem for f(x) = xÂ² – 4x – 3 in the interval [a, b], where a = 1, b = 4.

Solution:

f(x) = xÂ² – 4x – 3. Being a polynomial, it is continuous in the interval [1, 4] and derivable in (1, 4).

Question 5.

Verify Mean Value Theorem for f(x) = xÂ³ – 5xÂ² – 3x in the interval [a, b], where a = 1, b = 3. Find all c âˆˆ (1,3) for which f'(c) = 0.

Solution:

f(x) = xÂ³ – 5xÂ² – 3x.

It is a polynomial. Therefore, it is continuous in the interval [1, 3] and derivable in the interval (1, 3).

None of these values âˆˆ (1, 3).

Question 6.

Examine the applicability of Mean Value Theorem for all the three functions given in the question 2.

Solution:

(i) f(x) = [x] for x âˆˆ [5, 9],

f(x) = [x] is neither continuous in the interval [5, 9] nor differentiable in the interval (5, 9).

Hence, MVT is not applicable.

(ii) f(x) = [x], for x âˆˆ [- 2, 2].

Again f(x) = [x] is neither continuous in the

interval [- 2, 2] nor differentiable in the interval (- 2, 2).

Hence, MVT is not applicable.

(iii) f(x) = xÂ² – 1 for x âˆˆ [1, 2].

It is a polynomial. Therefore, it is continuous in the interval [1, 2] and differentiable in the interval (1, 2).

âˆ´ f’ (x) = 2x, f(1) = 1 – 1 = 0, f(2) = 4 – 1 = 3.

âˆ´ f'(c) = 2c.

By MVT, f'(c) = \(\frac { f(b) – f(c) }{ b-a }\)

2c = \(\frac { 3-0 }{ 2-1 }\)

âˆ´ c = \(\frac { 3 }{ 2 }\), which belongs to (1, 3).